I am trying to evaluate the integral below by differentiating through the integral. Let
$ F(a,b) :=\displaystyle\int_{0}^{\pi/2}\log\left(a^2\cos^2\left(x\right)+b^2\sin^2\left(x\right)\right)\,{\rm d}x$
For fixed $b$, and letting $g(t) = F(t,b)$ I am trying to justify
$g'(t) = \displaystyle\int_{0}^{\pi/2} \dfrac{2t\cos^2(x)}{t^2\cos^2\left(x\right)+b^2\sin^2\left(x\right)}\,{\rm d}x$
In order to justify this, I need to show
- $f(t,b) :=\log\left(t^2\cos^2\left(x\right)+b^2\sin^2\left(x\right)\right)\,{\rm d}x$ is integrable over $[0,\pi/2]$
- $\dfrac{\partial f}{\partial t}$ exists for all $t,x$
- There exists a dominating function such that $\dfrac{\partial f}{\partial t} \leq g$, a.e and $g \in L^1([0,\pi/2])$
I'm thinking 1) is straightforward since it's continuous, hence measurable, and since it's on a finite interval we can conclude that it is also integrable.
$\dfrac{\partial f}{\partial t} = \dfrac{2t\cos^2(x)}{a^2\cos^2\left(x\right)+b^2\sin^2\left(x\right)}$ so 2) is satisfied
In regards to finding a dominating function, can MVT be used here?
I'm also having some trouble evaluating the integral $g'(t)$, but $F(a,b) = \pi \log \left(\dfrac{a+b}{2}\right) \ \ (a,b>0)$ should be the solution.